14 CHAPTER L Manifolds and Vector Note that S(/) = Si if n = m = 1. So the theorem is proved in this case. We proceed by induction on m. So let m 1 and assume that the theorem is true for each smooth map P Q where dim(P) m. We prove that /(S2 \ S3) has measure 0. For each x G S2 \ S3 there is a linear differential operator P such that Pf(x) = 0 and -J-j(x) ^ 0 for some i,j . Let W be the set of all such points, for fixed P, i,j. It suffices to show that f(W) has measure 0. By assumption, 0 G E is a regular value for the function Pfl : W E. Therefore W is a smooth submanifold of dimension m 1 in E m . Clearly, S(/) fl W is contained in the set of all singular points of f\W : W Mn, and by induction we get that /((S 2 \ S3) H W) C / ( £ ( / ) n W) C / ( E ( / | W)) has measure 0. It remains to prove that /(S3) has measure 0. Every point of S3 has an open neighborhood W C U on which - ^ ^ 0 for some i, j . By shrinking W if necessary and applying diffeomorphisms, we may assume that E m _ 1 x E D Wx x W2 = W -t- E n _ 1 x E, (y, t) i- (y(j/, t), t). Clearly, (y, t) is a critical point for / if and only if y is a critical point for g( ,t).ThusZ(f)nW = {Jtew2&(9( , *)) x{t}). Since dim(Wi) = m - 1 , by induction we get that jjJn~1(g(Y i (g( ,t),t))) = 0, where At""1 is the Lebesque measure in R n _ 1 . By Fubini's theorem we get dt t€W 2 JV ^2 / 0rft = 0. D Jw2 1.19. Embeddings into m. Then M can be embedded into E n if (1) n = 2m + l (this is due to [228] see also [84, p. 55] or [26, p. 73]). (2) n = 2m (see [228]). (3) Conjecture (still unproved): The minimal n is n = 2m a(m) + 1, where a(ra) is the number of 1's in the dyadic expansion of m. There exists an immersion (see section (2)) M E n if (4) n = 2m (see [84]). (5) n = 2 m - l (see [228]). (6) Conjecture: The minimal n is n = 2m a(m). The article [34] claims to have proven this. The proof is believed to be incomplete.
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